Problem: $h(t) = 4t+5+2(g(t))$ $g(t) = t^{2}+3t$ $ h(g(-3)) = {?} $
Answer: First, let's solve for the value of the inner function, $g(-3)$ . Then we'll know what to plug into the outer function. $g(-3) = (-3)^{2}+(3)(-3)$ $g(-3) = 0$ Now we know that $g(-3) = 0$ . Let's solve for $h(g(-3))$ , which is $h(0)$ $h(0) = (4)(0)+5+2(g(0))$ To solve for the value of $h$ , we need to solve for the value of $g(0)$ $g(0) = 0^{2}+(3)(0)$ $g(0) = 0$ That means $h(0) = (4)(0)+5+(2)(0)$ $h(0) = 5$